Renormalons in Quantum Mechanics Cihan Pazarba s Bo gazi ci - - PowerPoint PPT Presentation

Renormalons in Quantum Mechanics

Cihan Pazarba¸ sı

Bo˘ gazi¸ ci University

based on 1906.07198 collobration with Dieter Van den Bleeken

06.09.2019, ICTP

Introduction

• Perturbative Series are generically divergent

f(λ) = fnλn (fn ∼ A−nn!)

• Borel summation −

→ Tame the divergence

Introduction

• Perturbative Series are generically divergent

f(λ) = fnλn (fn ∼ A−nn!)

• Borel summation −

→ Tame the divergence Im f(λ) = Res(e−s ˆ f(sλ))|s=A/λ

C+

Re(s) Im(s) Choosing C+ or C− results in complex conjugate results! − → Borel Ambiguity

n! = ∞ ds sne−s = ⇒ f(λ) = ∞ ds e−s ˆ f(sλ)

ˆ f(sλ) = ∞

n=0 fn n! (sλ)n = A A−sλ

C−

Introduction

• Perturbative Series are generically divergent

f(λ) = fnλn (fn ∼ A−nn!)

• Borel summation −

→ Tame the divergence

• Types of Divergences:
• Proliferation of number diagrams −

→ Tunneling

• Borel Ambiguity

B-ZJ

← − − − − − →

cancellation Instanton/WKB Action

Proliferation of Diagrams

(Ambigious)

Introduction

• Perturbative Series are generically divergent

f(λ) = fnλn (fn ∼ A−nn!)

• Borel summation −

→ Tame the divergence

• Types of Divergences:

IR-Region:

• dk2k2

ln µ

k2

n ∼

• dte−ttn ∼ n!

UV Region: dk2

k2

• ln k2

µ

n ∼

• dte−ttn ∼ n!
• Proliferation of number diagrams −

→ Tunneling

• Borel Ambiguity

B-ZJ

← − − − − − →

cancellation Instanton/WKB Action

• Remnant of renormalization process of loop
• Borel Ambiguity −

→ ?

Proliferation of Diagrams Loop Expansion

(Ambigious)

Introduction

Renormalon Problems:

• No resolution to Borel ambiguity!
• No proof/conjecture that the divergence will survive or cancel out!

Solution:

• Simplify the problem → Find the simplest toy model
• Try NR Quantum Mechanics?

Introduction

Renormalon Problems:

• No resolution to Borel ambiguity!
• No proof/conjecture that the divergence will survive or cancel out!

Solution:

• Simplify the problem → Find the simplest toy model
• Try NR Quantum Mechanics?
• Delta potentials for D ≥ 2 have UV divergence → Candidate for

renormalons in QM!

Introduction

Renormalon Problems:

• No resolution to Borel ambiguity!
• No proof/conjecture that the divergence will survive or cancel out!

Solution:

• Simplify the problem → Find the simplest toy model
• Try NR Quantum Mechanics?
• Delta potentials for D ≥ 2 have UV divergence → Candidate for

renormalons in QM!

Outline:

• Review of δ(2)(x) scattering.
• Construction of Renormalons in QM.
• Resolution of Renormalon ambiguity.

Renormalization in QM

Perturbative Scattering: S = S(0) − iT where T = V (GV )n T (1)

ren = λ2 l(Ef)

, l(Ef) =

1 4π log eiπz µ

T (1) = λ2

4π

∞

du2 u2

f −u2

Renormalization: Consider 2-body scattering with V = δ(2)(x, y) 1 Loop:

Renormalization in QM

Perturbative Scattering: S = S(0) − iT where T = V (GV )n T (1)

ren = λ2 l(Ef)

, l(Ef) =

1 4π log eiπz µ

T (1) = λ2

4π

∞

du2 u2

f −u2

Renormalization: Consider 2-body scattering with V = δ(2)(x, y) 1 Loop: Generalization to higher loops is immediate! N-1 Loop:

. . .

Renormalization:

. . . . . .

T (n)

ren = λn (l(Ef))n−1

(Only diagram) Easy to sum!

Renormalization in QM

Perturbative Scattering: S = S(0) − iT where T = V (GV )n

• NP Bound State: E = −µe4π/λ
• This mathces with exact solution
• No Renormalon Integral

T (1)

ren = λ2 l(Ef)

, l(Ef) =

1 4π log eiπz µ

T (1) = λ2

4π

∞

du2 u2

f −u2

Renormalization: Consider 2-body scattering with V = δ(2)(x, y) 1 Loop: Generalization to higher loops is immediate! N-1 Loop:

. . .

Renormalization: T = λ 1 − λ

4π(log Ef µ + iπ)

. . . . . .

T (n)

ren = λn (l(Ef))n−1

(Only diagram) Easy to sum!

Renormalons In Quantum Mechanics

. . .

. . .

4 Point Interaction

• Renormalon Loop Exists
• Particle number conserves
• But still complicated

H = P 2

2 + λ0 δ2(x, y) + V (x, y, z)

3D Model: ∗−⋆−⋆−. . .−⋆−∗

Renormalons In Quantum Mechanics

. . .

. . .

4 Point Interaction

• Renormalon Loop Exists
• Particle number conserves
• But still complicated

Write the problem with a combinations of background potentials

H = P 2

2 + λ0 δ2(x, y) + V (x, y, z)

3D Model:

T (n+3) = λ∗

• du2
• λ∗

4π

• log

Ef −u2

2

µ2

∗

• − iπ

n d2Q2

(2π)2 d2Qn+3 (2π)2 F (P2, Pn+3)

• u2=un+3

renormalon integral ∼ n!

∗−⋆−⋆−. . .−⋆−∗

Renormalons In Quantum Mechanics

. . .

. . .

4 Point Interaction

• Renormalon Loop Exists
• Particle number conserves
• But still complicated

Write the problem with a combinations of background potentials F (P2, Pn+3) =

˜ V (Pf −Pn+3) ˜ V (P2−Pi) (Ef −En+3)(Ef −E2)

• (x, y, z) → (Q, u)

H = P 2

2 + λ0 δ2(x, y) + V (x, y, z)

3D Model:

T (n+3) = λ∗

• du2
• λ∗

4π

• log

Ef −u2

2

µ2

∗

• − iπ

n d2Q2

(2π)2 d2Qn+3 (2π)2 F (P2, Pn+3)

• u2=un+3

renormalon integral ∼ n!

∗−⋆−⋆−. . .−⋆−∗

Renormalons In Quantum Mechanics

. . .

. . .

4 Point Interaction

• Renormalon Loop Exists
• Particle number conserves
• But still complicated

Write the problem with a combinations of background potentials

• (x, y, z) → (Q, u)
• This implies 3rd direction leads to the renormalon integral

H = P 2

2 + λ0 δ2(x, y) + V (x, y, z)

3D Model:

T (n+3) = λ∗

• du2
• λ∗

4π

• log

Ef −u2

2

µ2

∗

• − iπ

n d2Q2

(2π)2 d2Qn+3 (2π)2 F (P2, Pn+3)

• u2=un+3

renormalon integral ∼ n!

• Does it survive in the full expansion?
• Physical implication?

∗−⋆−⋆−. . .−⋆−∗

Renormalons In Quantum Mechanics

. . .

. . .

4 Point Interaction

• Renormalon Loop Exists
• Particle number conserves
• But still complicated

Write the problem with a combinations of background potentials

Questions

H = P 2

2 + λ0 δ2(x, y) + V (x, y, z)

V (x, y, z) = κδ(z cos θ − y sin θ)

Renormalons In Quantum Mechanics

θ → 0 = ⇒ Sθ=0 = S2dS1d − → No renormalon

∗ − ⋆ − . . . − ⋆ − ∗ ⋆ − . . . − ⋆

• a

−∗ − ⋆ − . . . − ⋆

• n−a

−∗ ∗ − ⋆ − . . . − ⋆

• n−a

−∗ − ⋆ − . . . − ⋆

• a

⋆ − . . . − ⋆

• a

−∗ − ⋆ − . . . − ⋆

• n−a−b

−∗ − ⋆ − . . . − ⋆

• b

H = P 2

2 + λ0 δ2(x, y) + V (x, y, z)

V (x, y, z) = κδ(z cos θ − y sin θ)

Renormalons In Quantum Mechanics

All Diagrams at Order κ2

θ → 0 = ⇒ Sθ=0 = S2dS1d − → No renormalon

Tn,2 = 9 2(cos θ log cos2 θ)2κ2µ− 3

2

λ 6π n (n − 3)!

Ambiguity: Im T2 = ∓ 9πi

2 (cos θ log cos2 θ)2κ2µ− 3

2 e− 6π λ λ

6π

2

θ = 0

− − − − → 0

∗ − ⋆ − . . . − ⋆ − ∗ ⋆ − . . . − ⋆

• a

−∗ − ⋆ − . . . − ⋆

• n−a

−∗ ∗ − ⋆ − . . . − ⋆

• n−a

−∗ − ⋆ − . . . − ⋆

• a

⋆ − . . . − ⋆

• a

−∗ − ⋆ − . . . − ⋆

• n−a−b

−∗ − ⋆ − . . . − ⋆

• b

+

H = P 2

2 + λ0 δ2(x, y) + V (x, y, z)

V (x, y, z) = κδ(z cos θ − y sin θ)

Renormalons In Quantum Mechanics

All Diagrams at Order κ2

Standart Borel summation θ → 0 = ⇒ Sθ=0 = S2dS1d − → No renormalon

Resolution to Renormalon Ambiguity

Ambiguity: Im T2 = ∓ 9πi

2 (cos θ log cos2 θ)2κ2µ− 3

2 e− 6π λ λ

6π

2

• No need to cancel imaginary part! ( T ∈ C)
• Is there a “natural” sign?

Resolution to Renormalon Ambiguity

Ambiguity: Im T2 = ∓ 9πi

2 (cos θ log cos2 θ)2κ2µ− 3

2 e− 6π λ λ

6π

2

• No need to cancel imaginary part! ( T ∈ C)

Take a step back − → T =

n

• dq f(q) l(p2

f − q2)n−1λn

Here there are two options:

• 1. First integrate, then sum (what we’ve done)
• 2. First sum, then integrate (Observe we have a geometric series!)

T =

• dq f(q)

λ 1−λ l(p2

f −q2)−

→ Pole requires an analytical continuation

• Is there a “natural” sign? (YES!)

Resolution to Renormalon Ambiguity

Ambiguity: Im T2 = ∓ 9πi

2 (cos θ log cos2 θ)2κ2µ− 3

2 e− 6π λ λ

6π

2

• No need to cancel imaginary part! ( T ∈ C)

Take a step back − → T =

n

• dq f(q) l(p2

f − q2)n−1λn

Here there are two options:

• 1. First integrate, then sum (what we’ve done)
• 2. First sum, then integrate (Observe we have a geometric series!)

T =

• dq f(q)

λ 1−λ l(p2

f −q2)−

→ Pole requires an analytical continuation Feynman prescription − → T =

• dq f(q)

λ 1−λ l(p2

f −q2+iǫ)

this is choosing forward time evolution! Im T > 0

(E → E+iǫ)

• Is there a “natural” sign? (YES!)

Resolution to Renormalon Ambiguity

Ambiguity: Im T2 = ∓ 9πi

2 (cos θ log cos2 θ)2κ2µ− 3

2 e− 6π λ λ

6π

2

• No need to cancel imaginary part! ( T ∈ C)

Take a step back − → T =

n

• dq f(q) l(p2

f − q2)n−1λn

Here there are two options:

• 1. First integrate, then sum (what we’ve done)
• 2. First sum, then integrate (Observe we have a geometric series!)

T =

• dq f(q)

λ 1−λ l(p2

f −q2)−

→ Pole requires an analytical continuation Feynman prescription − → T =

• dq f(q)

λ 1−λ l(p2

f −q2−iǫ)

this is choosing backward time evolution! Im T > 0 − → Same sign!

(E → E−iǫ)

• Is there a “natural” sign? (YES!)

Resolution to Renormalon Ambiguity

No Ambiguity: Im T2 = + 9πi

2 (cos θ log cos2 θ)2κ2µ− 3

2 e− 6π λ λ

6π

2 T = ∞

−∞ dq f(q) λ 1−λ l(p2

f −q2±iǫ) = 2

∞ dq feven(q)

λ 1−λ l(p2

f −q2±iǫ)

T =

n V (G0(E±iǫ)V )n

+iǫ −iǫ Defining time direction is equivalent to choosing the analytic half-plane of G(E)

• What is the difference with standart Borel summation?

Forward time contour Backward time contour

Resolution to Renormalon Ambiguity

No Ambiguity: Im T2 = + 9πi

2 (cos θ log cos2 θ)2κ2µ− 3

2 e− 6π λ λ

6π

2 T = ∞

−∞ dq f(q) λ 1−λ l(p2

f −q2±iǫ) = 2

∞ dq feven(q)

λ 1−λ l(p2

f −q2±iǫ)

T =

n V (G0(E±iǫ)V )n

+iǫ −iǫ Defining time direction is equivalent to choosing the analytic half-plane of G(E)

• What is the difference with standart Borel summation?

Forward time contour Backward time contour Borel Summation

• No distinction between C+

and C−

C+

Re(s) Im(s)

C−

Conclusion

• 1-particle non-relativistic QM can have renormalons that leads to

non-vanishing imaginary contribution.

• Renormalon singularity ←

→ NP Bound state

• No Renormalon ambiguity! The problem is resolved by a different

approach.

Conclusion

• 1-particle non-relativistic QM can have renormalons that leads to

non-vanishing imaginary contribution.

• Renormalon singularity ←

→ NP Bound state

• No Renormalon ambiguity! The problem is resolved by a different

approach.

Outlook

• Independent semi-classical interpretation? (Helps in QFT)
• New look at the Borel summation!